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# A firm is divided into four departments, each of which

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A firm is divided into four departments, each of which [#permalink]  12 Dec 2010, 22:42
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80% (02:48) correct 20% (02:15) wrong based on 5 sessions
A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department,what is the greatest number of distinct teams to which the project could be assigned?
(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

[Reveal] Spoiler:
I did receive an explanation but did not really understand. My way to comprehend this was as follows,

Since there are 3 people to be choosen (and none can be repeated from the same department);
the number of ways to choose the departments are 4C3 ways. For selecting a person from each department is 4C1 X 4C1 X 4C1 ways. So the total number of ways (I presume my calculation is leading to distinct teams) is
4C3 X 4C1X 4C1 X4C1 ways= 4^4 ways. Could anyone please validate my thinking or correct me. Thanks.
[Reveal] Spoiler: OA
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Re: Combinations-4 Department,4 People-Assign to a team of 3. [#permalink]  13 Dec 2010, 01:22
Expert's post
helloanupam wrote:
A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?
(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

I did receive an explanation but did not really understand. My way to comprehend this was as follows,

Since there are 3 people to be choosen (and none can be repeated from the same department);
the number of ways to choose the departments are 4C3 ways. For selecting a person from each department is 4C1 X 4C1 X 4C1 ways. So the total number of ways (I presume my calculation is leading to distinct teams) is
4C3 X 4C1X 4C1 X4C1 ways= 4^4 ways. Could anyone please validate my thinking or correct me. Thanks.

Basically we are asked to determine the # of different groups of 3 that can be formed so that no 2 members are from the same department.

$$C^3_4=4$$ - choosing which 3 departments out of 4 will provide with a member;
$$C^1_4*C^1_4*C^1_4=4^3$$ - choosing each member from the selected 3 departments;

So, total # of different groups will be: $$4*4^3=4^4$$.

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Re: Combinations-4 Department,4 People-Assign to a team of 3. [#permalink]  13 Dec 2010, 01:58
Thanks! Bunuel.
Your explanations have always been lucid.
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Re: Combinations-4 Department,4 People-Assign to a team of 3. [#permalink]  13 Dec 2010, 05:50
+1 for B

Thanks for explaining
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Choosing department members [#permalink]  15 Feb 2012, 07:06
A firm is divided into four departments, each of which contains
four people. If a project is to be assigned to a team of three
people, none of which can be from the same department,
what is the greatest number of distinct teams to which the
project could be assigned?

16

32

64

256

1032
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If the Q jogged your mind do Kudos me : )

Math Expert
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Kudos [?]: 40506 [0], given: 5540

Re: Choosing department members [#permalink]  15 Feb 2012, 07:17
Expert's post
rxs0005 wrote:
A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?

16

32

64

256

1032

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Re: A firm is divided into four departments, each of which [#permalink]  17 Feb 2012, 00:15
I'm still confused. How would this question be answered if I was using the slotting method?
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Re: A firm is divided into four departments, each of which [#permalink]  17 Feb 2012, 02:07
Expert's post
pitpat wrote:
I'm still confused. How would this question be answered if I was using the slotting method?

First member: 4*4=16 choices;
Second member: 4*3=12 choices (as one department already provided with a member);
Third member: 4*2=8 choices (as two departments already provided with members);

So, we have: 16*12*8.

Since the order of the members in a team does not matter (we don't actually have 1st, 2nd and 3rd members) then we should divide above by 3! to get rid of duplication: 16*12*8/3!=4^4.

As for the other approach. Consider this: $$C^3_4$$ is # of ways to choose which 3 departments will provide with an employee from 4 departments for the team. Now, since each chosen department can provide with 4 employees then total # of different teams will be $$C^3_4*4*4*4=4^4$$.

Hope it's clear.
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Re: A firm is divided into four departments, each of which   [#permalink] 17 Feb 2012, 02:07
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